如何使用 SQL 从一个 table 中找到另一个 table 中的第一个匹配结果？

Question

假设我有两个 tables:

Customer -
ID  | Name  | Etc
1   | One   |
2   | Two   |
3   | Three |
4   | Four  |
5   | Five  |
... | ...   |

Sales - 
Customer ID | Date | Amount
5           | 1/20 | 
5           | 3/19 | 5
3           | 8/19 | 3
7           | 3/20 | 13
3           | 9/20 | ...
1           | 3/20 | ...
1           | 1/20 | ...

我想做的是编写一个查询来查找每个客户的第一笔销售。我不确定该怎么做。我觉得这是按问题分组的，但我没有找到答案。

编辑：我觉得我的第一个数据 table 没有完全解释我的问题。（老实说，直到我编写解决方案时，我才意识到我的问题的这一方面）注意：每次销售有不止一个客户。

Sales - 
Sale ID | Customer ID | Date | Amount
1       | 5           | 1/20 | 
5       | 5           | 3/19 | 5
8       | 3           | 8/19 | 3
7       | 7           | 3/20 | 13
3       | 4           | 9/20 | ...
2       | 1           | 3/20 | ...
1       | 1           | 1/20 | ...

Answer 1

您可以使用子查询为每个客户的每个销售分配一个行号，按日期升序排序，然后 select 仅第一行：

SELECT "Customer ID", "Date", "Amount"
FROM (
  SELECT "Customer ID", "Date", "Amount",
         ROW_NUMBER() OVER (PARTITION BY "Customer ID" ORDER BY "Date") AS rn
  FROM Sales) s
WHERE rn = 1

Demo on SQLFiddle

Answer 2

Nick 的解决方案可能是最高效的，但如果您想在此处使用 GROUP BY，您可以这样做：

SELECT
    c.ID,
    c.Name,
    s1.Date,
    s1.Amount
FROM Customer c
INNER JOIN Sales s1 ON c.ID = s1."Customer ID"
INNER JOIN
(
    SELECT "Customer ID", MIN(Date) AS FirstSaleDate
    FROM Sales
    GROUP BY "Customer ID"
) s2
    ON s1."Customer ID" = s2."Customer ID" AND
       s1.Date = s2.FirstSaleDate
ORDER BY
    c.ID,
    c.Name;

Answer 3

在 Oracle 中，您可以使用 keep:

select customer_id, min(date) as first_sales_date,
       max(amount) keep (dense_rank first order by date asc) as first_amount
from sales
group by customer_id;

如何使用 SQL 从一个 table 中找到另一个 table 中的第一个匹配结果？

How to use SQL find first matching result from one table in another table?

sql

oracle

group-by

partition